At 27^{circ} C temperature,the time required | vedclass

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(B) For a $1^{st}$ order reaction,the rate constant $K$ is given by the formula: $K = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$.Given t

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$0.0693 \ s^{-1}$

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(B) For a $1^{st}$ order reaction,the rate constant $K$ is given by the formula: $K = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} ight)$.Given that the reaction is $75 \%$ complete,the remaining concentration $[A]_t = [A]_0 - 0.75 [A]_0 = 0.25 [A]_0$.Substituting the values: $K = \frac{2.303}{20} \log \left( \frac{[A]_0}{0.25 [A]_0} ight)$.$K = \frac{2.303}{20} \log (4) = \frac{2.303}{20} 	imes 0.6021 \approx \frac{1.386}{20} = 0.0693 \ s^{-1}$.Alternatively,$t_{75\%} = 2 	imes t_{1/2}$. Since $t_{1/2} = \frac{0.693}{K}$,then $20 = 2 	imes \frac{0.693}{K}$,which gives $K = \frac{1.386}{20} = 0.0693 \ s^{-1}$.

At $27^{\circ} C$ temperature,the time required for $75 \%$ completion of a first order reaction is $20 \ s$. What will be its rate constant?

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